Question: A person stands $12$ meters east of an intersection and watches a car driving away from the intersection to the north at $4$ meters per second. At a certain instant, the car is $9$ meters from the intersection. What is the rate of change of the distance between the car and the person at that instant (in meters per second)? Choose 1 answer: Choose 1 answer: (Choice A) A $15$ (Choice B) B $\dfrac{20}{3}$ (Choice C) C $4\sqrt{10}$ (Choice D) D $2.4$
Explanation: Setting up the math Let... $a(t)$ denote the distance between the car and the intersection at time $t$, $b$ denote the distance between the person and the intersection (which is always $12$ meters), and $c(t)$ denote the distance between the car and the person at time $t$. $a(t)$ $b$ $c(t)$ We are given that $a'(t)=4$ and $b=12$. We are also given that $a(t_0)=9$ for a specific time $t_0$. We want to find $c'(t_0)$. Relating the measures The measures relate to each other through the Pythagorean theorem: $\begin{aligned} \,[a(t)]^2+b^2&=[c(t)]^2 \\\\\\ [a(t)]^2+(12)^2&=[c(t)]^2 \end{aligned}$ We can differentiate both sides to find an expression for $c'(t)$ : $c'(t)=\dfrac{a(t)a'(t)}{c(t)}$ Using the information to solve In order to find $c'(t_0)$ we need to find $c(t_0)$. Using the Pythagorean theorem and the fact that $a(t_0)=9$ and $b=12$, we can find that $c(t_0)=15$. Now let's plug ${a(t_0)}={9}$, ${a'(t_0)}={4}$, and ${c(t_0)}={15}$ into the expression for $c'(t_0)$ : $\begin{aligned} c'(t_0)&=\dfrac{{a(t_0)}{a'(t_0)}}{{c(t_0)}} \\\\ &=\dfrac{({9})({4})}{({15})} \\\\ &=2.4 \end{aligned}$ In conclusion, the rate of change of the distance between the car and the person at that instant is $2.4$ meters per second. Since the rate of change is positive, we know that the distance is increasing.